There is a 6 metres wide alley. Both walls of the alley are perpendicular to the ground. Two ladders, one 10 metres long, the other 12 metres, are propped up from opposite corners to the adjacent wall, forming an X shape. All four feet of each ladder are firmly touching either the corner or the wall. The two ladders are also touching each other at the intersection of the X shape.

What is the distance from the point of intersection from the ground?

Let x be the horizontal component of the distance of the crossover point from the wall where the base of the 10-meter ladder rests. Let y be the height of the crossover point.

The top of the 10-meter ladder rests 8 meters above the ground (the triangle is a 3,4,5 rt triangle with dimensions doubled), and the top of the 12-meter ladder is 12√3 meters up (as it makes an angle of 60 degrees with the ground).

The equation of the 10-meter ladder is therefore
y=8x/6
and the 12-meter ladder is
y=(12(√3)/6)(6-x)
substituting and multiplying by 6,
8x = 12(√3)(6-x)
and
x=72(√3)/(8+12√3)
and y=8x/6 = 5.7765896... meters above the ground.

Posted by Charlie on 2003-04-21 09:01:01